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The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 × 107 V>m. The capacitor is to have a capacitance of 1.25 × 109 F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

Short Answer

Expert verified
The minimum plate area needed is approximately 0.0136 m².

Step by step solution

01

Understand the Problem

We need to find the minimum area of the plates of a parallel-plate capacitor given specific electrical properties: dielectric constant, dielectric strength, capacitance, and maximum potential difference.
02

Use the Capacitance Formula

Recall the formula for the capacitance of a parallel-plate capacitor: C=ε0εrAdwhere ε0 is the permittivity of free space (8.85×1012 F/m), εr is the relative permittivity (3.60), A is the area, and d is the distance between the plates. Here, C=1.25×109 F.
03

Apply Dielectric Strength Condition

The dielectric strength provides the maximum electric field (Emax) the material can withstand, given by Emax=1.60×107 V/m. The potential difference V is related to the electric field by V=Emaxd. Solve for d:d=VEmax=5500V1.60×107V/m3.44×104m.
04

Rearrange the Capacitance Formula for Area

We know C, ε0, εr, and d, and we can solve for A:A=Cdε0εr=1.25×109F3.44×104m8.85×1012F/m3.601.36×102m2.
05

Solution Conclusion

The calculation yields the minimum area of the capacitor plates necessary to meet the given requirements.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dielectric Strength
Dielectric strength is a critical property of insulating materials, especially ones used in capacitors. It refers to the maximum electric field that a material can withstand without breaking down or losing its insulating properties. In simple terms, it's the material's ability to resist an electric breakdown.
For a dielectric material, the dielectric strength is measured in volts per meter (V/m). A higher dielectric strength means that the material can endure a stronger electric field. This is essential in preventing breakdown in capacitors, ensuring the device functions correctly under high voltages.
In the context of capacitor design, knowing the dielectric strength helps determine how thin the material can be without risking failure. Choosing a material with a suitable dielectric strength ensures that the capacitor can operate safely at the intended voltage, as calculated in our exercise.
Parallel-Plate Capacitor
A parallel-plate capacitor is a straightforward type of capacitor that features two conducting plates separated by an insulating material called a dielectric. This setup is widely used due to its simplicity and effectiveness.
The capacitance of this kind of capacitor is determined by the area of the plates, the separation distance between the plates, and the properties of the dielectric in between. Larger plate areas and smaller separation distances increase capacitance, making for a more efficient energy storage device.
Parallel-plate capacitors are commonly used in electronic circuits and can be found in devices ranging from radios to laptops. Understanding the factors that influence their design, like the dielectric material and plate area, is key to optimizing their performance for various applications.
Capacitance Formula
The capacitance formula for a parallel-plate capacitor is: C=ε0εrAdwhere:
  • C is the capacitance in farads (F).
  • ε0 is the permittivity of free space (8.85×1012 F/m).
  • εr is the relative permittivity or dielectric constant of the material.
  • A is the area of the plates.
  • d is the separation distance between the plates.
This formula calculates how much electric charge a capacitor can store. The greater the capacitance, the more charge it can hold for a given voltage.
By rearranging the formula, we can find different variables, such as the minimum area of the plates required for a specific capacitance, as done in the exercise. This is crucial for designing capacitors that meet precise electrical criteria.

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Most popular questions from this chapter

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, what is the charge per unitlength λ on the capacitor?

Three capacitors having capacitances of 8.4, 8.4, and 4.2 μF are connected in series across a 36-V potential difference. (a) What is the charge on the 4.2-μF capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00× 106 V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

A parallel-plate air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 cm2, what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

An air capacitor is made by using two flat plates, each with area A, separated by a distance d. Then a metal slab having thickness a (less than d) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. P24.62). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C0 when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits a 0 and a d.

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